Problem: Factor the following expression: $x^2 + 3x - 28$
Explanation: When we factor a polynomial, we are basically reversing this process of multiplying linear expressions together: $ \begin{eqnarray} (x + a)(x + b) &=& xx &+& xb + ax &+& ab \\ \\ &=& x^2 &+& {(a + b)}x &+& {ab} \end{eqnarray} $ $ \begin{eqnarray} \hphantom{(x + a)(x + b) }&\hphantom{=}&\hphantom{ xx }&\hphantom{+}&\hphantom{ (a + b)x }&\hphantom{+}& \\ &=& x^2 & +& {3}x& & {-28} \end{eqnarray} $ The coefficient on the $x$ term is $3$ and the constant term is $-28$ , so to reverse the steps above, we need to find two numbers that add up to $3$ and multiply to $-28$ You can try out different factors of $-28$ to see if you can find two that satisfy both conditions. If you're stuck and can't think of any, you can also rewrite the conditions as a system of equations and try solving for $a$ and $b$ $ {a} + {b} = {3}$ $ {a} \times {b} = {-28}$ The two numbers $7$ and $-4$ satisfy both conditions: $ {7} + {-4} = {3} $ $ {7} \times {-4} = {-28} $ So we can factor the expression as: $(x + {7})(x {-4})$